University:
Massachusetts Institute of Technology
Course:
16.30 | Estimation and Control of Aerospace Systems
Academic year:

2021
Views:

469

Pages:

12
Author:

Nancy Reilly

Problem Set 3 2 poles, 1 zero. [Diagram of a root locus plot with a circle] or: (negative gain) Why is this a circle? Let: H(s) = (s + z) / ((s + p1)(s + p2)) with 0 ≤ p1 ≤ p2 < ∞ 1 + K * H(s) = 0 1 + K * (s + z) / ((s + p1)(s + p2)) = 0 (s + p1)(s + p2) + K(s + z) = 0 s^2 + (p1 + p2 + K)s + p1p2 + Kz = 0 Discriminant: Δ = (p1 + p2 + K)^2 - 4(p1p2 + Kz) = p1^2 + p2^2 + K^2 + 2p1p2 + 2Kp1 + 2Kp2 - 4p1p2 - 4Kz = p1^2 + p2^2 + K^2 - 2p1p2 + 2K(p1 + p2 - 2z) = (p1 - p2)^2 + 2K(p1 + p2 - 2z) + K^2 The range of K for which the discriminant is negative is found by finding the... #3 K(s) = K * (s + 2.5) / (s(s^2 + 2s + 2)(s + 5)(s + 1)) [Root locus plot] a) K = 0: s = -2.5, -5, -1 b) K = 1: s = -6.66, -0.54 ± 0.89i c) K = 5: s = -8.42, -1.29 ± 0.98i d) K = ∞: s = -2, -1, 0 e) K = 7: s = -6.66, -1.16 ± 1.1i f) jw-axis crossing: No jw-axis crossing g) Break-in: K = -11 h) Break-out: K = -0.92 #3 (cont'd) f) K(s) = K * (s + 1) / (s(s + 1)(s + 3)(s^2 + 2)) [Root locus plot] a) K = 0: s = -1, -3, ±√2i b) K = 1: s = -1.07, -0.52 ± 0.95i c) K = 5: s = -1.23, -0.5 ± 1.1i d) jw-axis crossing: No jw-axis crossing e) Break-in: K = -1 f) Break-out: K = 0.73 3 roots of (p1 - p2)^2 + 2K(p1 + p2 - 2z) + K^2 = 0. so. K = the discriminant of THAT equation is: reduced (p1 + p2 - 2z)^2 - (p1 - p2)^2 = (p1 + p2 - 2z - p1 + p2)(p1 + p2 - 2z + p1 - p2) = (2p1 - 2z)(2p2 - 2z) = 4(z - p1)(z - p2) > 0. Thus for K ∈ [2z - (p1 + p2) ± 2√((p1 + p2 - 2z)^2 - 4(z - p1)(z - p2)) / 2(z - p1)(z - p2)] The roots are complex of the first equation.

16.30 Problem Set 3 (Solutions)

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